Discrete Applied Mathematics on B-Colorings in Regular Graphs$

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Discrete Applied Mathematics 157 (2009) 1787–1793

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Discrete Applied Mathematics

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On b-colorings in regular graphs$

Mostafa Blidia a, Frédéric Maffray b,∗, Zoham Zemir a a Département de Mathématiques, Université de Blida, B.P. 270, Blida, Algeria b C.N.R.S, Laboratoire G-SCOP, 46 Avenue Félix Viallet, 38031 Grenoble Cedex, France article info a b s t r a c t

Article history: A b-coloring is a coloring of the vertices of a graph such that each color class contains a Received 30 July 2008 vertex that has a neighbor in all other color classes. El-Sahili and Kouider have conjectured Received in revised form 6 January 2009 that every d-regular graph with girth at least 5 has a b-coloring with d + 1 colors. We show Accepted 9 January 2009 that the Petersen graph infirms this conjecture, and we propose a new formulation of this Available online 10 February 2009 question and give a positive answer for small degree. © 2009 Elsevier B.V. All rights reserved. Keywords: Coloration b-coloring a-chromatic number b-chromatic number

1. Introduction

A proper coloring of a graph G = (V , E) is a mapping c from V to the set of positive integers (colors) such that any two adjacent vertices are mapped to different colors. Each set of vertices colored with one color is a stable set of vertices of G, so a coloring is a partition of V into stable sets. The smallest number k for which G admits a coloring with k colors is the chromatic number χ(G) of G. Many graph invariants related to colorings have been defined. Most of them try to minimize the number of colors used to color the vertices under some constraints. For some other invariants, it is meaningful to try to maximize this number. The b-chromatic number is such an example. When we try to color the vertices of a graph, a simple trick consists in starting from a coloring and trying to decrease the number of colors by reducing them in some way, for example by merging two color classes. This motivated the introduction of the achromatic number by Harary and Hedetniemi [6]: the achromatic number of a graph G is the largest integer k such that G admits a coloring with k colors for which there is an edge between any two color classes. Clearly, the process of merging suggested above is impossible if we have such a coloring. So the achromatic number is a measure of how hard it is to obtain a coloring with few colors. This inspired Irving and Manlove [8,13] to consider another procedure, which consists in trying to reduce the number of colors by transferring all vertices from one color class to other classes. A b-coloring is a proper coloring such that every color class i contains at least one vertex that has a neighbor in all the other classes. Any such vertex will be called a b-dominating vertex of color i. The b-chromatic number b(G) is the largest integer k such that G admits a b-coloring with k colors. For a graph G, and for any vertex v of G, the neighborhood of v is the set N(v) = {u ∈ V (G) | uv ∈ E(G)} and the degree of v is deg(v) = |N(v)|. Let ∆(G) be the maximum degree in G, and let m(G) be the largest integer k such that G has k vertices of degree at least k − 1. It is easy to see that every graph G satisfies b(G) ≤ m(G) ≤ ∆(G) + 1

$ This research was supported by the Algerian/French program CMEP/Tassili 05 MDU 639. ∗ Corresponding author.

(the first inequality follows from the fact that if G has any b-coloring with k colors then it has k vertices of degree at least k − 1; the second inequality follows from the definition of m(G)). Irving and Manlove [8,13] proved that every tree T has b-chromatic number b(T ) equal to either m(T ) or m(T ) − 1, and their proof is a polynomial-time algorithm that computes the value of b(T ). On the other hand, Kratochvíl, Tuza and Voigt [12] proved that it is NP-complete to decide if b(G) = m(G), even when restricted to the class of connected bipartite graphs such that m(G) = ∆(G) + 1. These NP-completeness results have incited researchers to establish bounds on the b-chromatic number in general or to find exact or approximate values for subclasses of graphs [1–4,7,9–11]. Here we focus on a recent problem concerning regular graphs. A graph G is d-regular if every vertex of G has degree equal to d. Note that every d-regular graph G satisfies m(G) = d + 1. The girth of G is the length of a shortest cycle in G. In [3], El-Sahili and Kouider pose the following question:

Question ([3]). Is it true that every d-regular graph G with girth g(G) ≥ 5 satisfies b(G) = d + 1? We observe that the Petersen graph offers a negative answer to this question. However, we also propose some positive results in the case where d ≤ 6 which suggest that the Petersen graph might be the only counterexample to the question.

2. Preliminary results

For integer k ≥ 3, we let Ck denote the cycle with k vertices.

Theorem 1. The Petersen graph has b-chromatic number 3.

Proof. Let G be the Petersen graph, with vertices v1, . . . , v5, w1, . . . , w5 such that v1, v2, v3, v4, v5 induce a 5-cycle in this order, w1, w3, w5, w2, w4 induce a 5-cycle in this order, and viwi is an edge for each i = 1,..., 5. Since χ(G) = 3, we have b(G) ≥ 3. Suppose that G admits a b-coloring with 4 colors. For j = 1,..., 4, let dj be a b-dominating vertex of color j, and let D = {d1,..., d4}. Note that each vertex dj must have exactly one neighbor of each of the three colors different from j. Suppose that D induces a stable set. So, up to symmetry, we can assume that D = {v1, v3, w4, w5}, with d1 = v1, d3 = v3, d2 = w4, d4 = w5. Without loss of generality, v2 has color 2, and so w2 does not have color 2. Since v1 is b-dominating, it must have a neighbor of color 4, which can only be w1, and a neighbor of color 3, which can only be v5. Since v3 is b- dominating, it must have a neighbor of color 4, which can only be v4, and a neighbor of color 1, which can only be w3. But then w5 cannot have a neighbor of color 2, a contradiction. So D does not induce a stable set. We may assume that d1, d2 are adjacent, and so, up to symmetry, that d1 = v1 and d2 = v2 (since all edges of the Petersen graph play the same role). Now, it is easy to see that, wherever d3 may be, there is a C5 of G that contains d1, d2, d3. So (since all C5’s of the Petersen graph play the same role), we can assume that d3 is one of v3, v4, v5. Up to symmetry, this leads to two cases.

Case 1: d3 = v3. Since v2 is b-dominating, it has a neighbor of color 4, which can only be w2. One of v4, v5, say v5, does not have color 4. Since v1 is b-dominating, it has a neighbor of color 4, which can only be w1, and a neighbor of color 3, which can only be v5. Since v3 is b-dominating, it has a neighbor of color 4, which can only be v4, and a neighbor of color 1, which can only be w3. Note that w5 can only have color 2. But now, the vertices of color 4 are w1, w2, v4 and no other vertex, and each of these three vertices has two neighbors of the same color, so none of them can be b-dominating, a contradiction.

Case 2: d3 = v4. Since v1 is b-dominating, it has a neighbor of color 3, which can only be w1, and a neighbor of color 4, which can only be v5. Likewise v2 has a neighbor of color 3, which can only be w2, and a neighbor of color 4, which can only be v3. But then v4 has two neighbors of color 4, so it cannot be b-dominating, a contradiction. We will need the following result, for which we give a new proof. Our proof is rather similar to the proof of Proposition 2 in [3]; we include it here for the sake of completeness and because it prepares the more complicated proof of Theorem 5.

Theorem 2 ([9]). Every d-regular graph G with girth g(G) ≥ 6 has a b-coloring with d + 1 colors.

Proof. Let x be a vertex of G, and let x1,..., xd be its neighbors. For each i = 1,..., d, let Ni = N(xi) \{x}. Then each Ni is a stable set, for otherwise G would contain a cycle of length three. Then any two Ni’s are disjoint, for otherwise G would contain a cycle of length four; and there is no edge between them, for otherwise G would contain a cycle of length five. We construct a coloring with d + 1 colors 0, 1,..., d as follows. Assign color 0 to x, color i to xi (i = 1,..., d), and assign to the vertices of Ni the colors from {1,..., d}\{i}, in a one-to-one fashion. Finally, color the remaining vertices in arbitrary order, assigning to each v a color from {0, 1,..., d} different from the colors already assigned to its neighbors. Clearly, we obtain a b-coloring with d + 1 colors in which the vertices x, x1,..., xd are b-dominating.

3. A new proof of El-Sahili and Kouider’s theorem

El-Sahili and Kouider [3] proved the following theorem.

Theorem 3 ([3]). If G is a d-regular graph with girth g(G) ≥ 5 and G contains no C6, then b(G) = d + 1. M. Blidia et al. / Discrete Applied Mathematics 157 (2009) 1787–1793 1789

We propose here a new proof of Theorem 3, using a classical theorem of Vizing on list coloring. Let L be a mapping that assigns to each vertex v of a graph G a set L(v) of admissible colors. An L-coloring is a coloring c of the vertices of G such that c(v) ∈ L(v) for every vertex v of G. If G admits an L-coloring it is said to be L-colorable. Given an integer k, G is k-list-colorable if it admits an L-coloring for every L such that |L(v)| ≥ k for every v ∈ V . The list-chromatic number χL(G) is the smallest integer k such that G is k-list-colorable.

Theorem 4 ([5,15]). Let G be a connected graph different from a complete graph and from an odd cycle. Then G is ∆-list-colorable, where ∆ is the maximum degree in G. Proof of Theorem 3. The Theorem holds trivially when d = 0 or 1. If d = 2, then G is a disjoint union of cycles, all of length 5 or more. Then b(G) = 3. Indeed, in order to obtain a b-coloring with 3 colors, it suffices to give colors 1, 2, 3, 1, 2 to five consecutive vertices in one cycle of G, and to color the rest of G with colors 1, 2, 3. Now we assume that d ≥ 3. Pick a vertex v of G and let its neighborhood be N(v) = {v1, v2, . . . , vd}. For each i = 1,..., d, let Ni = N(vi) \{v}, and let Gv be the subgraph of G induced by N1 ∪ · · · ∪ Nd. We make a few observations about Gv. Consider any vertex x in Gv; so x ∈ Ni for some i ∈ {1,..., d}. Vertex x has no neighbor in Ni, for otherwise G would contain a cycle of length three. Vertex x cannot have two neighbors y, z ∈ Nj (j 6= i), for otherwise x, y, vj, z would induce a cycle in G; and x cannot have two neighbors y ∈ Nj, z ∈ Nk (with i, j, k pairwise different), for otherwise x, y, vj, v, vk, z would induce a cycle in G. Thus, in Gv every vertex has degree at most 1. For i = 1,..., d, set Ei = {xy | x, y ∈ N(vi)\{v}, x 6= y}. Then let Hv be the graph obtained from Gv by adding to its edge set all the elements of E1 ∪ · · · ∪ Ed (so each Ni is a clique in Hv). It follows from the preceding observations that, in Hv, every vertex has degree at most d − 1. Moreover, if d ≥ 4, then the largest cliques in Hv are induced by the sets Ni (i = 1,..., d), which have size d − 1; and if d = 3 then the largest cliques in Hv have size 2. We claim that:

χL(Hv) ≤ d − 1. (1)

Consider the contrary case. Since the maximum degree in Hv is at most d − 1, by Theorem 4 we must have either (a) some component of Hv is a complete graph Kd, or (b) d − 1 = 2 and some component of Hv is an odd cycle. However, (a) is impossible because the largest cliques in Hv have size d − 1. So suppose that (b) holds. Then d = 3, so the graph Hv has six vertices, and so the only possible odd cycle in Hv is a C5, but then some vertex of this C5 has two neighbors in Gv, a contradiction. Thus (1) holds. We define a list assignment L on Hv as follows. If x is any vertex of Hv, we have x ∈ Ni for some i ∈ {1,..., d}, and we assign the list L(x) = {1,..., d}\{i} to x. Thus each vertex x of Hv satisfies |L(x)| = d − 1. By (1), there is an L-coloring c of Hv. Extend c to a coloring of G as follows. Give color 0 to v and color i to vi (i = 1,..., d); then color the remaining vertices in arbitrary order, giving to each vertex z a color from {0, 1,..., d} different from the colors already assigned to its neighbors. Then we obtain a coloring of G with d + 1 colors, and clearly the vertices v, v1, . . . , vd are b-dominating. Thus b(G) = d + 1.

4. When the degree is small

Theorem 5. Let G be a d-regular graph with girth g(G) ≥ 5, different from the Petersen graph, and with d ≤ 6. Then b(G) = d+1. Proof. We distinguish one case for each value of the degree. Case 1: d = 1. Then G is a matching and clearly b(G) = 2 = d + 1. Case 2: d = 2. Then G is a disjoint union of cycles, all of length at least 5. We can obtain a b-coloring with 3 colors by assigning colors 1, 2, 3, 1, 2 to five consecutive vertices in one cycle of G, and to color the rest of G with colors 1, 2, 3 greedily. So b(G) = 3. Now let d ≥ 3. We may assume that G contains a cycle of length 5, for otherwise the result follows from Theorem 2. Therefore in Cases 3, 4, and 5 below, we let x1,..., x5 be five vertices of G that induce a cycle C in this order.

Case 3: d = 3. For each i = 1,..., 5, let ui be the neighbor of xi that is not in the cycle C. First suppose that the edge uiui+2 exists for every i = 1,..., 5. Then the vertices x1,..., x5, u1,..., u5 induce the Petersen graph and form one connected component of G. Since G itself is not the Petersen graph, it must have another component Z. In that case, give color 1 to x1, x4, color 2 to x2, u5, color 3 to x3, u1, and color 4 to x5, u2, u3 and to some vertex z of Z, and give colors 1, 2, 3 to the neighbors of z. So x1, x2, x3, z are b-dominating vertices of colors 1, 2, 3, 4 respectively, and this coloring can be extended to a coloring of G with four colors in any greedy way. Now we can assume, up to symmetry, that u1u3 is not an edge of G. We construct a b-coloring with 4 colors such that x1, x2, x3, u2 are b-dominating vertices of colors 1, 2, 3, 4 respectively. To do this, we first give color 1 to x1, x4, color 2 to x2, color 3 to x3, x5, and color 4 to u1, u2, u3. Note that x1, x2, x3 are b-dominating vertices of colors 1, 2, 3 respectively. Now consider u2. Let a, b be the two neighbors of u2 different from x2. (Possibly {a, b} ∩ {u4, u5} 6= ∅.) Note that a and b are not adjacent to x1, x2, x3, for otherwise G would contain a cycle of length 3 or 4. Moreover, and for the same reason, each of a, b is adjacent to at most one of x4, x5; and if each of them is adjacent to one of x4, x5 then it is not to the same vertex; in other words the edge set between {a, b} and {x4, x5} is a matching of size at most two. So it is possible to give color 1 to one of a, b and color 3 to the other without having two adjacent vertices of the same color. Now u2 is a b-dominating vertex of color 4. Finally this coloring can be extended to a coloring of G with four colors in any greedy way. 1790 M. Blidia et al. / Discrete Applied Mathematics 157 (2009) 1787–1793

Fig. 1. When d = 5.

Case 4: d = 4. For each i = 1,..., 5, let Ai be the set of the two neighbors of xi that are not in C. Here all subscripts on the Ai’s are understood modulo 5 and from the set {1,..., 5}. Since G contains no cycle of length 3 or 4, it is easy to see that:

Ai is a stable set; (2)

Ai ∩ Aj = ∅ if i 6= j; (3)

There is no edge between Ai and Ai+1; (4)

We construct a b-coloring of G with five colors such that x1,..., x5 are b-dominating vertices of colors 1,..., 5 respectively, as follows. For each i = 1,..., 5, assign color i to xi and colors i + 2 and i + 3 (modulo 5) to the two vertices of Ai. Fact (4), and the fact that the two colors assigned to the vertices of Ai are different from the two colors assigned to Ai+2 resp. Ai+3, ensures that no two adjacent vertices in A1 ∪ · · · ∪ A5 receive the same color. Thus all vertices of A1,..., A5 have received a color, and each of x1,..., x5 has neighbors of all colors other than its own. Finally, since the uncolored vertices have degree 4, we can color them successively with one of the five colors, in any greedy way. Thus we obtain a b-coloring of G with five colors.

Case 5: d = 5. For each i = 1,..., 5, let Ai be the set of the three neighbors of xi that are not in C. All subscripts on A1,..., A5 are understood modulo 5 and from the set {1,..., 5}. Since G contains no cycle of length 3 or 4, it is easy to see that:

Ai is a stable set; (5)

Ai ∩ Aj = ∅ if i 6= j; (6)

There is no edge between Ai and Ai+1; (7)

Every vertex different from xi has at most one neighbor in Ai. (8)

For each i = 1,..., 5, we can find a neighbor si of xi such that the set S = {s1,..., s5} is a stable set, as follows. Pick any s1 ∈ A1, then s3 ∈ A3 \ N(s1), s5 ∈ A5 \ N(s3), s2 ∈ A2 \ N(s5), and s4 ∈ A4 \ (N(s1) ∪ N(s2)). Such vertices exist because of (7) and (8). It follows from this construction that S = {s1,..., s5} is indeed a stable set. We rename vertex s1 as x6. For i = 1,..., 5, let Bi = Ai \{si}; so |Bi| = 2. Let B6 = N(x6) \{x1}; so |B6| = 4. Since G contains no cycle of length 3 or 4, it is easy to see that:

B6 is a stable set; (9)

B6 ∩ (B1 ∪ B2 ∪ B5) = ∅; (10)

|B6 ∩ Bi| ≤ 1 for i ∈ {3, 4}; (11)

There is no edge between B6 and B1; (12)

Every vertex different from x6 has at most one neighbor in B6. (13) Note that condition (7) implies that:

There is no edge between Bi and Bi+1 (i ∈ {1,..., 5}, modulo 5); (14) and conditions (8) and (13) imply that:

The edges between Bi and Bj form a matching (i, j ∈ {1,..., 6}, i 6= j). (15)

We construct a b-coloring of G with six colors such that x1,..., x6 will be b-dominating vertices of colors 1,..., 6 respectively. We start by assigning color i to xi for each i = 1,..., 5 and color 6 to the vertices of S = {s1,..., s5}. Now we must find a way to assign colors i + 2 and i + 3 (modulo 5) to the two vertices of Bi, for each i = 1,..., 5, and colors 2, 3, 4, 5 to the four vertices of B6. We view this as a list-coloring problem, where each vertex of Bi (i = 1,..., 5) has a list of allowed colors Li = {i + 2, i + 3} and each vertex of B6 has a list of allowed colors L6 = {2, 3, 4, 5}. See Fig. 1. In that figure, each box represents a set Bi with its list Li; a line between two boxes means that there may be edges between the corresponding sets, subject to condition (15); and no line between two boxes illustrates conditions (12) and (14). During our coloring procedure, we will say that a vertex x loses a color j if this color must be removed from the list of allowed colors for x (because it has been assigned to a neighbor of x). M. Blidia et al. / Discrete Applied Mathematics 157 (2009) 1787–1793 1791

Recall from (11) that B6 may have one common vertex with any of B3, B4. Let the vertices of B6 be called a, b, c, d such that: if B6 ∩ B3 6= ∅, then a is the (unique) vertex in that intersection; and if B6 ∩ B4 6= ∅, then b is the (unique) vertex in that intersection. Assign colors 5, 2, 3, 4 to a, b, c, d respectively. By (14), and since the sets of colors we want to assign to Bi and Bi+2 (modulo 5) are disjoint, we can ignore the edges between any two such sets. Therefore we may color the sets B1,..., B5 independently from each other and no conflict will arise between any two sets. By (12), we can color the two vertices of B1 with colors 3 and 4. Because of the assignment in B6, and by (15), for each j ∈ {4, 5} at most one vertex of B2 loses color j, and that is a different vertex for each j. So it is possible to color the two vertices of B2 with colors 4 and 5. The same holds for B5 with colors 2 and 3. Now consider B3. If a ∈ B3, then a is a vertex of color 5 in B3 and the remaining vertex of B3 can be colored 1. If a 6∈ B3 then one vertex of B3 may lose color 5, but it is still possible to color the two vertices of B3 with colors 1 and 5. The same holds for B4 with colors 1, 2. Thus all vertices of B1,..., B6 have received a color, and each of x1,..., x6 has neighbors of all colors other than its own. Finally, since the uncolored vertices have degree 5, we can color them successively with one of the six colors, in any greedy way. Thus we obtain a b-coloring of G with six colors. Case 6: d = 6. The proof here uses similar arguments as in the case d = 5, but the situation is more complicated. We can assume that G contains a cycle of length 6, for otherwise the result follows from Theorem 3. Let x1,..., x6 be six vertices of G that induce a cycle in this order. For each i = 1,..., 6, let Ai = N(xi) \{xi−1, xi+1}; so |Ai| = 4. Here all subscripts on A1,..., A6 are understood modulo 6 and from the set {1,..., 6}. Since G contains no cycle of length 3 or 4, it is easy to see that:

Ai is a stable set; (16)

Ai ∩ Ai+1 = ∅ and Ai ∩ Ai+2 = ∅; (17)

|Ai ∩ Ai+3| ≤ 1; (18)

There is no edge between Ai and Ai+1; (19)

Every vertex different from xi has at most one neighbor in Ai. (20)

For each i = 1,..., 6, we find a neighbor si of xi such that the set S = {s1,..., s6} is a stable set, as follows:

– If A1 ∩ A4 6= ∅, let s1 and s4 be equal to the (unique) vertex in A1 ∩ A4. If A1 ∩ A4 = ∅, let s1 be any vertex in A1 and s4 be any vertex in A4 \ N(s1) (s4 exists by (20)). – If A3 ∩ A6 6= ∅, let s3 and s6 be equal to the (unique) vertex in A3 ∩ A6. Note that, by (19), this vertex is not adjacent to the vertices s1 and s4 found previously. If A3 ∩ A6 = ∅, let s3 be any vertex in A3 \ N(s1) and s6 be any vertex in A6 \ (N(s4) ∪ N(s3)) (vertices s3 and s6 exist by (20)). – If A5 ∩ A2 6= ∅, let s5 and s2 be equal to the (unique) vertex in A5 ∩ A2. Note that, by (19), this vertex is not adjacent to any of the vertices s1, s3, s4, s6 found previously. If A5 ∩ A2 = ∅, then, by (20), there are at least two vertices in A5 \ (N(s1) ∪ N(s3)) and at least two vertices in A2 \ (N(s4) ∪ N(s6)); and by (20) again, among these four vertices there are non-adjacent vertices s5 ∈ A5 and s2 ∈ A2.

It follows from this construction that the set S = {s1,..., s6} is a stable set. We rename vertex s1 as x7. For i = 1,..., 6, let Bi = Ai \{si}; so |Bi| = 3. Note that B1,..., B6 are pairwise disjoint by (17), (18) and the definition of {s1,..., s6}. Let B7 = N(x7) \{x1}; so |B7| = 5. Since G contains no cycle of length 3 or 4, it is easy to see that:

B7 is a stable set; (21)

B7 ∩ (B1 ∪ B2 ∪ B6) = ∅; (22)

|B7 ∩ Bi| ≤ 1 for i ∈ {3, 4, 5}; (23)

There is no edge between B7 and B1; (24)

Every vertex different from x7 has at most one neighbor in B7. (25) Note that condition (19) implies that:

There is no edge between Bi and Bi+1 (i ∈ {1,..., 6}, modulo 6); (26) and conditions (20) and (25) imply that:

The edges between Bi and Bj form a matching (i, j ∈ {1,..., 7}, i 6= j). (27) In particular:

If sets X ⊆ Bi and Y ⊆ Bj are such that |X| > |Y |, then some vertex of X has no neighbor in Y . (28)

We construct a b-coloring of G with seven colors such that x1,..., x7 will be b-dominating vertices of colors 1,..., 7 respectively. We start by assigning color i to xi for each i = 1,..., 6 and color 7 to the vertices of S = {s1,..., s6}. Now we must find a way to assign colors i + 2, i + 3, i + 4 (modulo 6) to the three vertices of Bi, for each i = 1,..., 6, and colors 2, 3, 4, 5, 6 to the five vertices of B7. We view this as a list-coloring problem, where each vertex of Bi (i = 1,..., 6) has a list 1792 M. Blidia et al. / Discrete Applied Mathematics 157 (2009) 1787–1793

Fig. 2. When d = 6.

of allowed colors Li = {i + 2, i + 3, i + 4} and each vertex of B7 has a list of allowed colors L7 = {2, 3, 4, 5, 6}. See Fig. 2. In that figure, each box represents a set Bi with its list Li; a line between two boxes means that there may be edges between the corresponding sets, subject to condition (27); and no line between two boxes illustrates conditions (24) and (26). During our coloring procedure, we will say that a vertex x loses a color j if this color must be removed from the list of allowed colors for x.

Recall from (23) that B7 may have one common vertex with any of B3, B4, B5. Up to symmetry, we may assume that |B7 ∩ B3| ≥ |B7 ∩ B5|, in other words, if B7 intersects one of B3, B5 then it intersects B3. Define vertices a, b, c of B7 as follows, where we distinguish two cases:

Case (i): B7 ∩ B4 6= ∅. Let a be the (unique) vertex in B7 ∩ B4. Then, if B7 ∩ B3 6= ∅, let b be the (unique) vertex in B7 ∩ B3 (note that b has no neighbor in B4 by (19)); else, let b be a vertex in B7 \{a} that has no neighbor in B4 (such a vertex exists by (28)). Finally, if B7 ∩ B5 6= ∅, let c be the vertex in B7 ∩ B5; else, let c be any vertex in B7 \{a, b}.

Case (ii): B7 ∩ B4 = ∅. If B7 ∩ B3 6= ∅, let b be the (unique) vertex in this intersection (note that b has no neighbor in B4 by (19)); else, let b be a vertex in B7 that has no neighbor in B4 (such a vertex exists by (28)). Then, if B7 ∩ B5 6= ∅, let c be the vertex in this intersection; else, let c be any vertex in B7 \{b}. Finally, let a be any vertex in B7 \{b, c}. Note that in all cases vertices a, b, c are well defined and different since B3, B4, B5 are pairwise disjoint as mentioned above; and b has no neighbor in B4. Then a vertex d ∈ B7 is chosen as follows:

If b ∈ B3, c ∈ B5, a has a neighbor v5 ∈ B5 and v5 has a neighbor v3 ∈ B3, then choose d in B7 \{a, b, c} and not adjacent to v3 (such a vertex exists by (28)); else let d be any vertex in B7 \{a, b, c}. Finally let e be the remaining vertex of B7. Assign colors 2, 6, 3, 5, 4 to a, b, c, d, e respectively. Pick a vertex f2 ∈ B2 not adjacent to e and a vertex f6 ∈ B6 not adjacent to e or f2; such vertices exist by (20). Assign color 4 to f2 and f6.

Let i ∈ {1,..., 6}. By (19) there is no edge between Bi and Bi−1 ∪ Bi+1. Moreover, the sets of colors we want to assign to Bi and to Bi+3 are disjoint, so we can ignore the edges between these two sets. Therefore we can color B1 ∪ B3 ∪ B5 and B2 ∪ B4 ∪ B6 independently of each other with no conflict between the two sets. Let us consider B2, B4, B6. Because of the assignment in B7, and by (20), for each j ∈ {5, 6} at most one vertex of B2 \{f2} loses color j (a different vertex for each j). So it is possible to assign colors 5 and 6 to the two vertices of B2 \{f2}. Likewise, for each k ∈ {2, 3} at most one vertex of B6 \{f6} loses color k (a different vertex for each k), so it is possible to assign colors 2 and 3 to the two vertices of B6 \{f6}. Call t the vertex of B6 that receives color 2; so t is not adjacent to a. We are left with coloring the vertices of B4. First suppose that a ∈ B4 (case (i)). Then a is a vertex of color 2 in B4 (recall that a, t are not adjacent), and the two vertices of B4 \{a} lose color 2. Because of the assignment in B2 ∪ B6, and by (20), at most one vertex of B4 can lose a color (color 6), and by the choice of b no other vertex of B4 can lose color 6. So it is possible to assign colors 1 and 6 to the two vertices of B4 \{a}. Now suppose that a 6∈ B4 (case (ii)). Because of the assignment in B7 ∪ B2 ∪ B6, and by (20), at most two vertices of B4 lose color 2, and by the choice of b at most one loses color 6. So it is possible to assign colors 1, 2, 6 to the three vertices of B4. Thus, in either case all vertices of B2 ∪ B4 ∪ B6 ∪ B7 have received a color, and each of x2, x4, x6, x7 has neighbors of all colors other than its own. Now we deal with B3, B5. First suppose that c ∈ B5, which implies b ∈ B3 since |B7 ∩ B3| ≥ |B7 ∩ B5|. Then the vertices of B3 \{b} lose color 6 and the vertices of B5 \{c} lose color 3. Because of the assignment in B7, at most one vertex v3 of B3 \{b} loses a color (color 5) and at most one vertex v5 of B5 \{c} loses a color (color 2). We assign color 1 to v3 and v5. Note that, by the choice of d, we may assume that v3 and v5 are not adjacent. Then we assign color 5 to the third vertex of B3 and color 2 to the third vertex of B5. Now suppose that c 6∈ B5. Because of the assignment in B7, for each j ∈ {2, 3} at most one vertex of B5 loses color j (a different vertex for each j), so some vertex w5 ∈ B5 loses no color. If b ∈ B3, then the vertices of B3 \{b} lose color 6 and, because of the assignment in B7, at most one vertex v3 of B3 \{b} loses a color (color 5). So we assign color 1 to v3 and color 5 to the remaining vertex of B3. Because of this assignment in B3, at most one vertex of B5 loses color 1. So it is possible to assign colors 1, 2, 3 to the three vertices of B5. If b 6∈ B3, then for each j ∈ {5, 6} at most one vertex of B3 loses color j (a different vertex for each j), so some vertex w3 ∈ B3 loses no color. By (20), among the four vertices of (B3 ∪ B5) \{w3, w5}, there are two non-adjacent vertices, one in B3 and one in B5, to which we assign color 1. Then it is possible to assign color 5 and 6 to the remaining vertices of B3 and colors 2 and 3 to the remaining vertices of B5. M. Blidia et al. / Discrete Applied Mathematics 157 (2009) 1787–1793 1793

Now we deal with B1. Recall that (22) and (24) hold. Because of the assignment in B3, at most one vertex of B1 loses a color (color 5) and because of the assignment in B5 at most one vertex of B1 (possibly the same vertex) loses a color (color 3). So it is possible to color the three vertices of B1 with the colors 3, 4, 5. Thus all vertices of B1 ∪ B3 ∪ B5 have received a color, and each of x1, x3, x5 has neighbors of all colors other than its own. Finally, since the uncolored vertices have degree 6, we can color them successively with one of the seven colors, in any greedy way. Thus we obtain a b-coloring of G with seven colors. The proof above illustrates a technique which can probably not be extended to the general case. Indeed we tried to make a similar proof for graphs with d = 7, but the case analysis seems to become inextricable.

Remark. In view of the case d = 7, we may consider the so-called Hoffman–Singleton graph HS, which is the smallest 7- regular graph with girth at least 5. This graph is famous for many interesting properties related to its highly symmetric structure; see [14,16]. It is natural to suspect that HS might be a counterexample to El-Sahili and Kouider’s question. However, the b-chromatic number of HS is 8 (it is not hard to construct a b-coloring of HS with eight colors such that one vertex and its neighbors are b-vertices of colors 1,..., 8). In conclusion we propose the following reformulation of El-Sahili and Kouider’s question:

Conjecture 1. Every d-regular graph with girth at least 5, different from the Petersen graph, has a b-coloring with d + 1 colors.

Acknowledgments

We thank András Gyárfás for the remark that precedes Conjecture 1. We are grateful to the referees for their careful reading of the manuscript and useful comments and suggestions.

References

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